knitr::opts_chunk$set(echo = TRUE)
library(pagedown)
setwd("/Users/dunk/Classes/MTH463/Homework 4")
pagedown::chrome_print(input = "Homework-4.html",
                      output = "Homework4.pdf")

$$
\begin{aligned}

\end{aligned}
$$

$$
\begin{aligned}
\frac{d}{dx}e^{2x}&=\boxed{?} \\
&= \boxed{2e^{2x}}
\end{aligned}
$$

---
title: 'Homework #X'
author: "Name Here"
date: "`r format(Sys.time(), '%d %B, %Y')`"
output: 
  prettydoc::html_pretty:
    theme: leonids # Other options are cayman, tactile, architect, leonids, hpstr
    highlight: vignette
    math: katex #Instead of mathjax so it works offline
documentclass: book
classoption:
  - twocolumn
  - landscape
papersize: a5
linestretch: 1.5
fontsize: 12pt
links-as-notes: true
header-includes:
  - \usepackage{indentfirst}
---

Problem 1

I check that f(x) is a probability density function by finding if the cumulative distribution function is 1

\[ \begin{aligned} \int_{-\infty}^{\infty}f(x)dx&=\int_{-2}^{\infty}\frac{8dx}{x^3} \\ &=\Bigg[\frac{8x^{-2}}{-2}\Bigg]_{2}^{\infty} \\ &= \boxed{1} \end{aligned} \] So we know that f(x) is indeed a probability density function.

To find \(P(X>5)\)

\[ \begin{aligned} P(X>5)&=\int_{5}^{\infty}\frac{8dx}{x^3} \\ &= \Bigg[\frac{8x^{-2}}{-2}\Bigg]_{5}^{\infty} \\ &= \boxed{\frac{4}{25}} \end{aligned} \]

To find \(E[X]\) we use

\[ \begin{aligned} E[X]&=\int_{-\infty}^{\infty}xf(x)dx \\ &=\int_{2}^{\infty}\frac{8dx}{x^2} \\ &=\Bigg[\frac{8x^{-1}}{-1}\Bigg]_{2}^{\infty} \\ &=\boxed{4} \end{aligned} \]

Problem 2

I find c by integrating knowing the probability density function is equal to 1

\[ \begin{aligned} 1&=\int_{-\infty}^{\infty}f(x)dx \\ &= \int_{1}^{2}c(x-1)^4dx \\ &= \Bigg[\int_{1}^{2}\frac{c(x-1)^5}{5}\Bigg]_1^2 \\ &=\frac{c}{5} \\ \text{therefore, } &\boxed{ c=5} \end{aligned} \]

To find \(E[X]\) we have

\[ \begin{aligned} E[X]&=\int_{-\infty}^{\infty}xf(x)dx \\ &=\int_{1}^{2}5x(x-1)^4dx \\ &=\int_{1}^{2}5((x-1)+1)(x-1)^4dx \\ &=\int_{1}^{2}5(x-1)^5dx+\int_{1}^{2}5(x-1)^4dx \\ &=\Bigg[\frac{5(x-1)^6}{6}\Bigg]_1^2+1 \\ &= \frac{5}{6}+1 \\ &= \boxed{\frac{11}{6}} \end{aligned} \]

Problem 3

\[ \begin{aligned} 1&= \int_{-\infty}^{\infty}f(x)dx \\ &= \int_0^1(ax^2+bx)dx \\ &= \Bigg[ax^3\frac{1}{3}+bx^2\frac{1}{2}\Bigg] \\ &= \frac{1}{3}a+\frac{1}{2}b \end{aligned} \]

\[ \begin{aligned} 0.75&=E(X) \\ &=\int_{-\infty}^{\infty}xf(x)dx \\ &=\int_{0}^{1}(ax^3+bx^2)dx \\ &= \Bigg[a\frac{x^4}{4}+b\frac{x^3}{3}\Bigg]^1_0 \\ &=\frac{1}{4}a+\frac{1}{3}b \end{aligned} \]

So we have \(1 = \frac{1}{3}a+\frac{1}{2}b\) and \(0.75= \frac{1}{4}a+\frac{1}{3}b\) which reduces to \(b=0, a = 3\).

Therefore,

\[ \begin{aligned} E(X^2)&=\int_{-\infty}^{\infty}x^2f(x)dx \\ &=\int_{1}^{0}3x^4dx \\ &= \frac{3x^5}{5}\Bigg|_0^1 \\ &= \boxed{\frac{3}{5}} \end{aligned} \]

and,

\[ \begin{aligned} Var(X)&= E(X^2)-E(X)^2 \\ &= \frac{3}{5} - {0.75}^2 \\ &= \boxed{0.0375} \end{aligned} \]

Problem 4

\[ \begin{aligned} P(1<X<3)&=\int_1^3F(x)dx \\ &=\int_{-\infty}^3F(x)dx- \int_{-\infty}^1F(x)dx \\ &=F(3)-F(1) \\ &= \frac{15}{6}-\frac{3}{4} \\ &= \boxed{\frac{3}{16}} \end{aligned} \]

Problem 5

Given \(4x^2+4xY-Y+6=0\) with \(\lambda=3\), the roots of Y (\(\frac{-4Y \pm \sqrt{16Y^2+16(Y-6)}}{8}\)) will only be positive iff \(16Y^2+16(Y-6) \ge 0\). So we need to find

\[ \begin{aligned} P(16Y^2+16(Y-6) \ge 0)&=P(Y^2+Y-6 \ge 0) \\ &=P((Y+3)*(Y-2) \ge 0) \\ &=P(Y \le -3) + P(Y \ge 2) \\ &=0+e^{-2\lambda} \\ &= \boxed{e^{-6}} \end{aligned} \]

Problem 6

\[ \begin{aligned} \tau(\alpha + 1)&=\int_0^{\infty}e^{-y}y^{\alpha}dy \\ &= \int_0^{\infty}(-e^{-y})'y^{\alpha}dy \\ &=(-e^{-y}y^\alpha)\Bigg|_0^{\infty}-\int_0^{\infty}e^{-y}(y^{\alpha})'dy \\ &=0 + \int_0^{\infty}e^{-y}y^{\alpha-1}dy \\ &=\alpha \tau(\alpha) \\ \tau(1)&=\int_0^{\infty}e^{-y}dy \\ &=1=0! \end{aligned} \]

Therefore \(\tau(2)=1*\tau(1)=1!\)

Problem 7

\[ \begin{aligned} E[X^k]&=\int_0^{\infty}(\frac{y}{\lambda})^ke^{-y}dy \\ &=\frac{1}{\lambda^k}\int_0^{\infty}e^{-y}y^{(k+1)-1}dy \\ &=\frac{\tau(k+1)}{\lambda^k} \\ &=\frac{k!}{\lambda^k} \end{aligned} \]

Problem 8

\[ \begin{aligned} E[e^{-x}]&=\frac{1}{\tau(\alpha)}\int_0^{\infty}\lambda e^{-x}e^{-\lambda x}(\lambda x)^{\alpha -1}dx \\ &=\frac{\lambda}{\tau(\alpha)}\int_0^{\infty}e^{-(\lambda + 1) x}(x)^{\alpha -1}dx && \text{Let } y = (\lambda + 1)x \\ &=\frac{\lambda}{\tau(\alpha)}\int_0^{\infty}e^{-y}\frac{y^{\alpha - 1}}{(\lambda + 1)^{\alpha - 1}}\frac{1}{\lambda + 1}dy \\ &=\Bigg(\frac{\lambda}{\lambda+1}\Bigg)^\alpha\frac{1}{\tau(\alpha)}\int_0^{\infty}e^{-y}y^{\alpha-1}dy \\ &=\Bigg(\frac{\lambda}{\lambda+1}\Bigg)^\alpha \end{aligned} \]

Problem 9

We have \(t>0\), \(f(t)=\lambda e^{-\lambda t}\), \(F(t)=1-e^{-\lambda t}\).

\[ \begin{aligned} h(t)&=\frac{f(t)}{1-F(t)} \\ &=\frac{\lambda e^{-\lambda t}}{1-(1-e^{-\lambda t})} \\ &= \lambda \end{aligned} \]

The memorylessness property therefore makes it such that the “hazard rate” is constant at all times